| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.3 This is a standard M3 mechanics question requiring chain rule differentiation (a = v dv/dx), separation of variables, and integration. Part (a) is routine calculus, parts (b) and (c) involve straightforward integration and algebraic manipulation. While it requires multiple techniques, these are all standard M3 methods with no novel insight needed. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = v\frac{dv}{dx} = \frac{-x}{x+1} \times \frac{-2}{(x+1)^2} = \frac{-4}{(x+1)^3}\) | M2 A2 | |
| \(\frac{dx}{dt} = \frac{-x}{x+1} \therefore \int (x+1)\,dx = \int 2\,dt\) | M2 | |
| \(\frac{1}{2}x^2 + x = 2t + c\) | A1 | |
| \(t = 0, x = 0 \therefore c = 0\) so \(x^2 + 2x = 4t\) | M1 | |
| \(t = T, x = d\) \(\therefore d^2 + 2d = 4T\) | A1 | |
| \(t = T + 9, x = d + 4 \therefore (d + 4)^2 + 2(d + 4) = 4(T + 9)\) | M1 | |
| combining, \(d^2 + 8d + 16 + 2d + 8 = d^2 + 2d + 36\) | M1 | |
| giving \(8d = 12\) so \(d = 1.5\) | A1 | |
| \((1.5)^2 + 2(1.5) = 4T\) giving \(T = \frac{21}{16}\) or \(1.3125\) s | M1 A1 | (14) |
| Total | (75) |
| $a = v\frac{dv}{dx} = \frac{-x}{x+1} \times \frac{-2}{(x+1)^2} = \frac{-4}{(x+1)^3}$ | M2 A2 | |
| $\frac{dx}{dt} = \frac{-x}{x+1} \therefore \int (x+1)\,dx = \int 2\,dt$ | M2 | |
| $\frac{1}{2}x^2 + x = 2t + c$ | A1 | |
| $t = 0, x = 0 \therefore c = 0$ so $x^2 + 2x = 4t$ | M1 | |
| $t = T, x = d$ $\therefore d^2 + 2d = 4T$ | A1 | |
| $t = T + 9, x = d + 4 \therefore (d + 4)^2 + 2(d + 4) = 4(T + 9)$ | M1 | |
| combining, $d^2 + 8d + 16 + 2d + 8 = d^2 + 2d + 36$ | M1 | |
| giving $8d = 12$ so $d = 1.5$ | A1 | |
| $(1.5)^2 + 2(1.5) = 4T$ giving $T = \frac{21}{16}$ or $1.3125$ s | M1 A1 | (14) |
**Total** | (75) |7. A particle is travelling along the $x$-axis. At time $t = 0$, the particle is at $O$ and it travels such that its velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at a distance $x$ metres from $O$ is given by
$$v = \frac { 2 } { x + 1 }$$
The acceleration of the particle is $a \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = \frac { - 4 } { ( x + 1 ) ^ { 3 } }$.\\
(4 marks)
The points $A$ and $B$ lie on the $x$-axis. Given that the particle travels $d$ metres from $O$ to $A$ in $T$ seconds and 4 metres from $A$ to $B$ in 9 seconds,
\item show that $d = 1.5$,
\item find $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [14]}}